I want to prove that |HK| = |H||K| / |H ∩ K| using group actions. First, I let H and K be subgroups of G and the set S be defined as the set of left cosets of K in G. I define a group action by a * (xK) = axK where and .

by considering the orbit of the element K in S, i have that the orbit of K is the set of all s = a*K = aK, such that . but this is just the set theoretic product HK, so the orbit of K is equal to HK. From the orbit stabilizer theorem, the number of elements in the orbit of K is equal to the number of left cosets of the stabilizer of K. The stabilizer of K is the set of all a such that a*K = aK = K and . but this implies that and so . so the stabilizer of K is the set .

now the number of left cosets of the stabilizer of K is |H|/|H ∩ K|. therefore the orbit stabilizer theorem says that |HK| = |H|/|H ∩ K|, but i seem to be missing a factor of |K| somewhere and going through the problem I can't seem to find where it must come from. Can someone help show me what went wrong? thanks.

December 2nd 2011, 06:30 PM

Drexel28

Re: prove that |HK| = |H||K| / |H ∩ K| (using group actions)

Quote:

Originally Posted by oblixps

I want to prove that |HK| = |H||K| / |H ∩ K| using group actions. First, I let H and K be subgroups of G and the set S be defined as the set of left cosets of K in G. I define a group action by a * (xK) = axK where and .

by considering the orbit of the element K in S, i have that the orbit of K is the set of all s = a*K = aK, such that . but this is just the set theoretic product HK, so the orbit of K is equal to HK. From the orbit stabilizer theorem, the number of elements in the orbit of K is equal to the number of left cosets of the stabilizer of K. The stabilizer of K is the set of all a such that a*K = aK = K and . but this implies that and so . so the stabilizer of K is the set .

now the number of left cosets of the stabilizer of K is |H|/|H ∩ K|. therefore the orbit stabilizer theorem says that |HK| = |H|/|H ∩ K|, but i seem to be missing a factor of |K| somewhere and going through the problem I can't seem to find where it must come from. Can someone help show me what went wrong? thanks.

The problem is that your orbit is not . The elements of your orbit are which is not equal to .One's a set of cosets one is a set of group elements(don't worry, I had to stare at it for five minutes too. I was like "this guy just disproved math". haha, we all do stupid things, no worries).

I would suggest letting act on by . Then, and . Clearly then by the orbit stabilizer theorem .

I want to prove that |HK| = |H||K| / |H ∩ K| using group actions. First, I let H and K be subgroups of G and the set S be defined as the set of left cosets of K in G. I define a group action by a * (xK) = axK where and .

by considering the orbit of the element K in S, i have that the orbit of K is the set of all s = a*K = aK, such that . but this is just the set theoretic product HK, so the orbit of K is equal to HK. From the orbit stabilizer theorem, the number of elements in the orbit of K is equal to the number of left cosets of the stabilizer of K. The stabilizer of K is the set of all a such that a*K = aK = K and . but this implies that and so . so the stabilizer of K is the set .

now the number of left cosets of the stabilizer of K is |H|/|H ∩ K|. therefore the orbit stabilizer theorem says that |HK| = |H|/|H ∩ K|, but i seem to be missing a factor of |K| somewhere and going through the problem I can't seem to find where it must come from. Can someone help show me what went wrong? thanks.

K is not an a single element it is a set which has l K l elements
you can check this thread look into post #2

May 19th 2012, 12:52 AM

math819

Re: prove that |HK| = |H||K| / |H ∩ K| (using group actions)

I want to ask do HK is a subgroup of G ??
If yes, DO we need to assume one of the H, K is normal subgroup of G to ensure HK is normal in G ??
If HK is not a subgroup of G, will the equality become Inequality??

May 19th 2012, 01:32 PM

Deveno

Re: prove that |HK| = |H||K| / |H ∩ K| (using group actions)

HK is usually NOT a subgroup of G. if one of H,K is normal in G, then HK will be a subgroup:

let's say K is normal:

then (hk)(h'k') = h(kh')k', and since K is normal, Kh' = h'K, so kh' = k"h', for some k" in K, so

h(kh')k' = h(h'k")k' = (hh')(k"k'), which is in HK, so HK is closed under multiplication.

note that (hk)-1 = k-1h-1.

since Kh-1 = h-1K, k-1h-1 = h-1k', for some k' in K, so:

(hk)-1 = h-1k', which is in HK (since h-1 is in H, because H is a subgroup).

thus HK contains all inverses of elements hk in HK, and is thus a subgroup of G.

IF HK is a subgroup, there is still no guarantee that HK will be a NORMAL subgroup.